CHAPTER 3EXERCISE WITH SOLUTION1. The following table shows a partial probability distribution for the MRA Companys projected profits (in thousands of dollars) for the first year of operation (the negative value denotes a loss):

-1000.10

00.20

500.30

1000.25

1500.10

200

a) Find the missing value of f (200). What is your interpretation of this value?b) What is the probability that MRA will be profitable?c) What is the probability that MRA will make at least $100,000?Answer 1a) The missing value of is 0.05. b) The MRA will be profitable with the probability of 0.70. c) 0.40 is the probability that the MAR will make at least $100,000.

2. Data were collected on the number of operating rooms in use at Tampa General Hospital over a 20-day period. On 3 of the days only one operating room was used; on 5 days, two were used; on 8 days, three were used; and on 4 days all four rooms were used. a) Use the relative frequency approach to construct a probability distribution for the number of operating rooms in use on any given day.b) Draw a graph of the probability distribution.c) Show that your probability distribution satisfies the requirements for a valid discrete probability distribution.Answer 2a) No. of operating roomsXNo. of daysProbability distribution

13

25

38

44

Total201

c)The given data is satisfying the requirements of discrete probability distribution which are as follows.i. ii. 3. Brandon Lang is a creative entrepreneur who has developed a novelty soap item called Jackpot to target consumers with a gambling habit. Inside each bar of Jackpot shower soap is a single rolled-up bill of U.S. currency. The currency (rolled up and sealed in shrink-wrap) is appropriately inserted into the soap mixture prior to the cutting and stamping procedure. The distribution of paper currency (per 1000 bars of soap) is given in the following table.Distribution of Paper Currency Prizes

Bill DenominationNumber of Bills

$1520

$5260

$2070

$10120

$5029

$1001

Total1000

a) What is the expected amount of money in single bar of Jackpot soap?b) What is the standard deviation of the money in single bar of Jackpot soap?c) How many bars of soap would a customer have to buy so that, on average, he or she has purchased three bars containing a $50 or $100 bill?d) If a customer buys 8 bars of soap, what is the probability that at least one of these bars contains a bill of $20 or larger?Answer 3a) Expected amount = Distribution of Paper Currency Prizes

Bill Denomination

Number of Bills

$10.520.52

$50.261.3

$100.121.2

$200.071.4

$500.291.45

$1000.0010.1

Total15.97

Expected amount = 5.97b) Standard deviation =

Distribution of Paper Currency Prizes

Bill Denomination

Number of Bills

Expected value

$10.520.522412.48

$50.261.30.940.244

$100.121.216.241.94

$200.071.4196.8413.77

$500.291.451938.6456.22

$1000.0010.18841.648.84

Total15.97 93.494

Standard Deviation =

.Let Y = number of bars that contain a $50 or $100 bill. Observe that Y is a binomial random variable, where n = numbers of bars purchased and p = probability of a bar containing a $50 or $100 bill = (29/1000) + (1/1000) = .03. E[Y] = n p = n 0.03 = 3, thus n = 100. The customer needs to buy 100 bars of soap to have, on average, three bars with a $50 or $100 bill.

d.Let W = number of bars that contain a $20 or $50 or $100 bill. Observe that W is a binomial random variable, where n = 8 and p = (70 + 29 + 1 / 1000) = 0.10. P(W 1) = 1 P(W = 0) = 1 - .4305 = .5695.

4. The demand for Carolina Industries product varies greatly from month to month. Based on the past two years of data, the following probability distribution shows the companys monthly demand:

Unit DemandProbability

3000.20

4000.30

5000.35

6000.15

a) If the company places monthly orders equal to the expected value of the monthly demand, what should Carolinas monthly order quantity be for this product?b) Assume that each unit demanded generates $70 in revenue and that each unit ordered costs $50. How much will the company gain or lose in a month if it places an order based on your answer to part (a) and the actual demand for the item is 300 units?c) What are the variance and standard deviation for the number of units demanded?Answer 4a) Expected value = Unit Demand

Probability

3000.2060

4000.30120

5000.35175

6000.1590

Total1445

Thus the Carolinas monthly order quantity will be 445.

b) When the units are 445 at a price of $70 the revenue is 70*445= 31150 and the cost is 50*445=$22250. So in this scenario company is gaining (Profit= Revenue Cost) $8900.sWhen the units are 300 at a price of $70 the revenue is 70*300 = $21000 and the cost is 50*300= $15000. So the company is gaining (Profit= Revenue Cost) $6000.c) I. Variance = Unit Demand

Probability

Expected value

3000.2060210254205

4000.301202025608

5000.3517530251059

6000.1590240253604

Total14459476

Variance = 9476.II. Standard deviation = = Standard deviation = 97.5. The J. R. Ryland Computer Company is considering a plant expansion that will enable the company to begin production of a new computer product. The companys president must determine whether to make the expansion a medium- or large-scale project. The demand for the new product involves an uncertainty, which for planning purposes may be low demand, medium demand, or high demand. The probability estimates for the demands are 0.20, 0.50, and 0.30, respectively. Letting x indicate the annual profit in $1000s, the firms planners developed profit forecasts for the medium- and large-scale expansion projects.

Medium scale Expansion profits Large scale Expansion profits

Demand

Low500.2000.20

Medium1500.501000.50

High2000.303000.30

a) Compute the expected value for the profit associated with the two expansion alternatives. Which decision is preferred for the objective of maximizing the expected profit?b) Compute the variance for the profit associated with the two expansion alternatives. Which decision is preferred for the objective of minimizing the risk or uncertainty?Answer 5a) Expected value = Medium scale Expansion profitsLarge scale Expansion profits

Expected value

Expected value

DemandLow500.201000.200

Medium1500.50751000.5050

High2000.30603000.3090

Total1=1451=140

So we have found the medium scale expansion will maximize the profits of the company with $1, 45,000.b) Variance =

Medium scale Expansion profitsLarge scale Expansion profits

Expected value

Expected value

DemandLow500.2010180500.2003920

Medium1500.5075121000.5050800

High2000.30609073000.30907680

Total1=145=27241=140=12400

To minimize the risk and uncertainty the medium scale expansion profits variance is more preferable that is 2724.

6. A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January 28, 2008). The survey found that 23% of the respondents have boycotted goods for ethical reasons.a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?c) In a sample of ten British citizens, what is the probability that none have boycotted goods for ethical reasons?Answer 6a) Probability of 2 boycotted goodsBinomial probability distribution = n = 6 = no. of trialsp = 0.23 = probability of success on one trialx = 2 = number of successes in n trials Sf(x) = probability of x successes in n trials. b) Probability of at least 2 boycotted goods. n = 6 = no. of trialsp = 0.23 = probability of success on one trialx = 2 = number of successes in n trials f(x) = probability of x successes in n trials. We can find this probability by subtracting the probability of 0 and 1 boycotted goods from 1 (whole probability).

c) None have boycotted goods.n = 10 = no. of trialsp = 0.23 = probability of success on one trialx = 0 = number of successes in n trials f(x) = probability of x successes in n trials.

7. When a new machine is functioning properly, only 3% of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. a) Describe the conditions under which this situation would be a binomial experiment?b) How many experimental outcomes yield one defect?c) Compute the probabilities associated with finding no defects, one defect, and two defects?Answer 7a) The trials are identical and they are just defective or non defective. Probabilities of each trial are not changing on repetition (Independent). So, we can say that this information is meeting conditions of binomial probability distribution.b) Yield one defect

n = 2 = no. of trialsp = 0.03 = probability of defective itemx = 1 = number of defective item in n trials f(x) = probability of x successes in n trials.

c) Probabilities I. With no defectX= 0

II. With one defect.X=1

III. With two defectX= 2

8. Military radar and missile detection systems are designed to warn a country of enemy attacks. A reliability question deals with the ability of the detection system to identify an attack and issue the warning. Assume that a particular detection system has a 0.90 probability of detecting a missile attack. Answer the following questions using the binomial probability distribution:a) What is the probability that one detection system will detect an attack?b) If two detection systems are installed in the same area and operate independently, what is the probability that at least one of the systems will detect the attack?c) If three systems are installed, what is the probability that at least one of the systems will detect the attack?d) Would you recommend that multiple detection systems be operated? Explain.Answer 8a) n = 1 = no. of detection systemp = 0.90 = probability of detecting missile attackx = 1 = number of detection system in n trials f(x) = probability of x successes in n trials.

b) n = 2 = no. of detection systemp = 0.90 = probability of detecting missile attackx = 1 = number of detection system in n trials f(x) = Probability of x successes in n trials.

c) n = 3 = no. of detection systemp = 0.90 = probability of detecting missile attackx = 1 = number of detection system in n trials f(x) = probability of x successes in n trials. We can find this probability by subtracting the probability of 0 detection system from 1 (whole probability).

d) There is not as such big difference between single and multiple detection system. In my point of view 1 system is enough but to gain perfection or maximum security we can use multiple systems.9. Consider a Poisson probability distribution with 2 as the average number of occurrences per time perioda) Write the appropriate Poisson probability function?b) What is the average number of occurrences in three time periods?c) Write the appropriate Poisson probability function to determine the probability of x occurrences in three time periods?d) Find the probability of two occurrences in one time period?e) Find the probability of six occurrences in three time periods?f) Find the probability of five occurrences in two time periods?Answer 9a) Poisson probability function =

b) Average no. of occurrence

So the average no. of occurrence is 6 for 3 time periods.

c) Probability functions in 3 time period.

d) 2 occurrence in 1 time period= 2 = mean or average number of occurrences in an intervale = 2.71828x = 2 = number of occurrences in the intervalf(x)= probability of x occurrences in the intervale) Probability of 6 occurrences in 3 time period.= 2 = mean or average number of occurrences in an intervale = 2.71828x = 6 = number of occurrences in the intervalf(x)= probability of x occurrences in the interval

f) Probability of 5 occurrences in 2 time period.= 2 = mean or average number of occurrences in an intervale = 2.71828x = 5 = number of occurrences in the intervalf(x)= probability of x occurrences in the interval

10. Telephone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways.a) Find the probability of receiving 3 calls in a 5-minute interval?b) Find the probability of receiving 10 calls in 15 minutes?c) Suppose that no calls are currently on hold. If the agent takes 5 minutes to complete processing the current call, how many callers do you expect to be waiting by that time?d) If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted? Answer 10a) 3 calls in 5 minutes interval = 48 = mean or average calls in one houre = 2.71828x = 3 = number of occurrences in the intervalf(x)= probability of x occurrences in the interval

As the data is given in minutes format so we will convert no. of calls in minute format.= 4 = mean or average calls in 5 minutes. x = 3 = number of calls receivedb) 10 calls in 15 minutes.= 12 = mean or average calls in 15 minutes.x = 10 = number of calls received

c) = 48 (5 / 60) = 4 I expect 4 callers to be waiting after 5 minutes.

The probability none will be waiting after 5 minutes is .0183.11. More than 50 million guests stayed at bed and breakfasts (B & Bs) last year. The website for the Bed and Breakfast Inns of North America, which averages approximately seven visitors per minute, enables many B & Bs to attract guests without waiting years to be mentioned in guidebooks (Time, September 2001).a) What is the probability of no website visitors in a 1 minute period?b) What is the probability of two or more website visitors in a 1 minute period?c) What is the probability of one or more website visitors in a 30 second period?d) What is the probability of five or more website visitors in a 1 minute period?Answer 11= 7 = mean or average visitors in one minutee = 2.71828x = number of visitors in the intervalf(x)= probability of x occurrences in the intervala) No visitor in one minute period

b) 2 visitors in one minuteWe can calculate the probability by subtracting the probability of 2 from 1.

c) 1 visitors in 30 seconds= 3.5 = mean or average visitors in 30 seconds.We can calculate the probability by subtracting the probability of 1 from 1.

d) 5 visitors in one minute

12. Airline passengers arrive randomly and independently at the passenger screening facility at a major international airport. The mean arrival rate is 10 passengers per minute.a) What is the probability of no arrivals in a 1 minute period?b) What is the probability of 3 or fewer arrivals in a 1 minute period?c) What is the probability of no arrivals in a 15 second period?d) What is the probability of at least 1 arrival in a 15 second period?Answer 12= 10 = mean or average passenger in one minute e = 2.71828 x = number of visitors in the intervalf(x)= probability of x occurrences in the intervala) No arrivals in one minute.

b) 3 arrivals in a minute.

c) No arrivals in 15 seconds.As the requirement is in seconds then well convert the average in seconds format.

d) 1 arrivals in 15 seconds.

13. A random variable x is uniformly distributed between 1.0 and 1.5. a) Show the graph of the probability density function.

b) Find P(x = 1.25).P(x = 1.25) = 0. The probability of any single point is zero since the area under the curve above any single point is zero.c) Find P(1.00